If $AB = AC$, the result is obvious due to symmetry (both statements are equivalent to $P$ being the midpoint of $BC$). WLOG assume $AB < AC$. Let $D$ be the intersection of the line $BC$ and the tangent at $A$ to the circumcircle $(ABC)$.

If $\overline{MP}$ and $\overline{NP}$ are equal, then $MN$ is parallel to the tangent at $P$ to $\Gamma$, which is $BC$. It follows that
$$\angle DAM = \angle DAB = \angle ACB = \angle ANM,$$
which implies $DA$ is tangent to $\Gamma$. Therefore, $\Gamma$ is tangent to $(ABC)$ at $A$.

Conversely, if $\Gamma$ and $(ABC)$ are tangent at $A$, then $DA$ is tangent to $\Gamma$. Therefore, we have
$$\angle ANM = \angle DAM = \angle DAB = \angle ACB,$$
which implies $MN \parallel BC$. This shows $P$ is the midpoint of $\overline{MN}$ as desired.